KinexiaLEARN PHYSICS

Radioactive Activity — Banana Equivalent Dose (Worked Example)

Hard
Problem

A banana contains about 0.50 g of potassium, of which 0.012% is radioactive K-40 (half-life 1.25 × 10⁹ years, molar mass ≈ 40 g/mol). What is the banana's activity?

Given

  • K-40 mass = 6.0 × 10⁻⁸ kg

  • t½ = 1.25 × 10⁹ y

  • M = 0.040 kg/mol

Find

The activity A.

Principle

Tiny K-40 fraction, enormous half-life — A = λN quantifies the famous 'banana dose'.

Formula

Solution

  1. t1/2=1.25×109×3.156×107=3.94×1016 st_{1/2} = 1.25 \times 10^9 \times 3.156 \times 10^7 = 3.94 \times 10^{16}\ \mathrm{s}

  2. λ=0.693/3.94×1016=1.76×10−17 s−1\lambda = 0.693/3.94 \times 10^{16} = 1.76 \times 10^{-17}\ \mathrm{s^{-1}}

  3. N=(6.0×10−8/0.040)(6.022×1023)=9.03×1017N = (6.0 \times 10^{-8}/0.040)(6.022 \times 10^{23}) = 9.03 \times 10^{17}

  4. A=(1.76×10−17)(9.03×1017)=15.9 BqA = (1.76 \times 10^{-17})(9.03 \times 10^{17}) = 15.9\ \mathrm{Bq}

15.9 Bq

Unit check

s⁻¹ × atom count = s⁻¹ = Bq. The units confirm a decay rate, as required.

What this means physically

~16 Bq — the origin of the 'banana equivalent dose' meme. Your own body contains ~4000 Bq of K-40; one banana is utterly negligible radiation.

Common mistake

Using 0.50 g as the K-40 mass. Only 0.012% of potassium is K-40 — read the isotopic fraction, not the element mass.